Problem: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{x^3 + 7x^2 - 8x}{-2x^3 - 16x^2 + 18x}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {x(x^2 + 7x - 8)} {-2x(x^2 + 8x - 9)} $ $ r = -\dfrac{x}{2x} \cdot \dfrac{x^2 + 7x - 8}{x^2 + 8x - 9} $ Simplify: $ r = - \dfrac{1}{2} \cdot \dfrac{x^2 + 7x - 8}{x^2 + 8x - 9}$ Since we are dividing by $x$ , we must remember that $x \neq 0$ Next factor the numerator and denominator. $ r = - \dfrac{1}{2} \cdot \dfrac{(x - 1)(x + 8)}{(x - 1)(x + 9)}$ Assuming $x \neq 1$ , we can cancel the $x - 1$ $ r = - \dfrac{1}{2} \cdot \dfrac{x + 8}{x + 9}$ Therefore: $ r = \dfrac{ -x - 8 }{ 2(x + 9)}$, $x \neq 1$, $x \neq 0$